Divide $(x^{3} - 8)$ by $(x - 2)$.

Divide $(x^{3} - 8)$ by $(x - 2)$.
  1. $x^{2} + 2x + 4$
  2. $x^{2} - 2x + 4$
  3. $x^{2} + 2x - 4$
  4. $x^{2} - 4$

Solution

Use $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$. Here $x^{3}-8 = (x-2)(x^{2}+2x+4)$. Quotient $= x^{2}+2x+4$.

Asked in: IMO

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